A Boston Red Sox baseball player catches a ball of mass m that is moving toward him with speed v.While bringing the ball to rest, his hand moves back a distance d.Assuming constant deceleration, the horizontal force exerted on the ball by his hand is:

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The work done by the player's hand in bringing the ball to rest is given by the work-energy theorem: W = ΔK, where W is the work done by the force exerted by the hand, ΔK is the change in the kinetic energy of the ball, and we assume that all the work done by the hand is transferred to the ball. The initial kinetic energy of the ball is K1 = (1/2)mv^2, and the final kinetic energy is K2 = 0. The distance over which the force is applied is d, and the deceleration is a. Therefore, we have: W = ΔK = K2 - K1 = -(1/2)mv^2 The work done by the hand is also given by the formula: W = Fd where F is the force exerted by the hand, and d is the distance over which it is applied. Therefore, we have: Fd = -(1/2)mv^2 Solving for F, we get: F = -(1/2)mv^2/d = -mv^2/(2d) Note that the negative sign just indicates that the force is opposite to the direction of motion of the ball. To find the magnitude of the force, we take the absolute value: |F| = mv^2/(2d) Therefore, the horizontal force exerted on the ball by the player's hand is given by: Answer: E) mv²/(2d)

A Boston Red Sox Baseball Player Catches a Ball of Mass